Calculus questio ..........?

Question

if lim x approaching inifinity f(x) = 0 and lim x approaching infinity
g(x) = 0, does lim x approaching infinity f(x) / g(x) exist ? Explain with examples.

Answer 1

not necessarily.

f(x) = 1/x
g(x) = 1/x^2

f(x)/g(x) = (1/x) / (1/x^2) =x^2/x = x
so
lim x->0 f(x)/g(x) = lim x-> infinity x = infinity

now,
if f(x)= 1/x^2
and g(x) = 1/x
f(x)/g(x) = (1/x^2)/ (1/x) = x/x^2 = 1/x
so
lim x-> infinity f(x)/g(x) = lim x->infinity 1/x =0 .

Answer 2

It may or may not exist. In particular, it may exist and be finite, it may exist and be infinite, or it may fail to exist at all.

Here's an example with a finite limit:

f(x) = cx, g(x) = x, [x→0]lim f(x)/g(x) = c

And one with an infinite limit:

f(x) = |x|, g(x) = x², [x→0]lim f(x)/g(x) = ∞

And one with no limit at all:

f(x) = x, g(x) = x², [x→0]lim f(x)/g(x) does not exist, because [x→0+] lim f(x)/g(x) = ∞ and [x→0-]lim f(x)/g(x) = -∞

Here's one where even the left and right-hand limits don't exist:

f(x) = x sin (1/x), g(x) = x, [x→0]lim f(x)/g(x) does not exist, nor do either of the left- or right-hand limits (since f(x)/g(x) = sin (1/x), which oscillates back and forth between 1 and -1 infinitely many times in any interval surrounding 0). Note that [x→0]lim f(x) is in fact equal to zero, since sin (1/x) is always between 1 and -1, so x sin (1/x) is always between -x and x, and both of these quantities approach zero as x→0.

Answer 3

ok my friend....
look...
If lim x F(x)= 0 and
lim x g(x)= 0

then :

limx f(x) / lim xg(x) = 0/0.....right???...
then : 0/0 ...---> does not exist because there is an asymptote in the graph at 0.....take your calculator and type...0/0....it does not give you anything.....does it???

Answer 4

Your question doesn't have enough information to answer.

If g(x) isn't approching zero much faster than f(x) then the limit you asked about could exist. Otherwise it can not.

By much faster I don't mean orders of magnatude faster, I mean exponentialy faster.

Answer 5

f(x)=1/x
g(x)=1/x

limx->inf of f(x)/g(x)=1